1. ERT 207 ANALYTICAL CHEMISTRY ACIDS AND BASES THEORIES ACID BASE EQUILIBRIA IN WATER pH SCALE 27 Jan 2011 (MISS NOORULNAJWA DIYANA YAACOB) 1

2. Brønsted Acids and Bases A Brønsted acid is a species that donates a proton. (a proton donor). HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) A Conjugate base is what remains of the acid after the donation of a proton. A Brønsted base is a species that accepts a proton. (a proton acceptor). NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) A Conjugate acid is a newly formed protonated species. 2 acid conjugate base base conjugate acid hydronium ion

3. Brønsted Acids and Bases 3 Any reaction, using the Brønsted theory, involves both an acid and a base. Water is an amphoteric species which can act as a Brønsted acid or as a Brønsted base.

4. Brønsted Acids and Bases 4

5. Exercise: Identify acids, conjugate acids, bases and conjugate bases in the reactions below: (a) H 2 SO 4 (aq) + H 2 O(l) HSO 4 ‒ (aq) + H 3 O + (aq) (b) NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) Exercise: What is the conjugate base of H 2 PO 4 ‒ ? H 2 PO 4 ‒ (aq) + H 2 O(l) HPO 4 2‒ (aq) + H 3 O + (aq) 5 acid conjugate basebaseconjugate acid acid conjugate basebaseconjugate acid conjugate base

6. Lewis Acids and Bases The Brønsted base is the species that accepts a proton. The Brønsted acid is the species that donates a proton. A Lewis base is a species that donates a pair of electrons. A Lewis acid is a species that accepts a pair of electrons. 6 H+H+ N H H H + O H + H+H+ ‒ O H H H N H H H + Lewis base Lewis acid Lewis base Lewis acid

7. Lewis Acids and Bases Lewis acid-base reactions are more general than Brønsted acid-base reactions. Lewis acid-base reactions result in the formation of coordinate covalent bonds. 7

8. ARRHENIUS THEORY-H + AND OHˉ An acid is any substance that ionizes (partially or completely) in water to give hydrogen ions Means associate with solvent to give hydronium ions HA +H 2 O ↔ H 3 O+ + Aˉ

9. A base ionizes in water to give hydroxyl ions. Weak (partially ionizes) bases ionizes as follows: B + H 2 O ⇋ BH + + OHˉ For strong bases e.g.: (NaOH) M(OH) n →M n+ + nOHˉ this theory is obviously restricted to water as a solvent only

10. Franklin’s theory Recognizes the ionization of a solvent to give a cation and an anion E.g: 2H 2 O ⇋ H 3 O + + OHˉ 2NH 3 ⇋ NH 4 + NH 2 ¯ Acid = a solute that yields the cation of the solvent Base = a solute that yields the anion of the solvent Eg of strong acid is NH 4 Cl in liquid ammonia Eg of strong base is NaNH 2 in ammonia

11. ACID-BASE EQUILIBRIA IN WATER When acid or base is dissolved in water, it will dissociate Amount of ionization depends on the strength of the acid Strong electrolyte completely dissociated Weak electrolyte dissociated partially

12. Example: a)Hydrochloric acid (strong acid) a)Acetic acid (weak acid) Equilibrium constant is the thermodynamic activity constant

13. H + will be used instead of H 3 O + for simplification Molar concentration will be replaced by [] K a and K w are molar equilibrium constants [H + ]=1.0×10 -7 M = [OH - ]

14. The Acid-Base Properties of Water Water is a very weak electrolyte. It undergoes ionization to a very small extent. H 2 O (l) H + (aq) + OH ‒ (aq) 2H 2 O (l) H 3 O + (aq) + OH ‒ (aq) The equilibrium expression of water autoionization is: K w = [H + ] [OH ‒ ] or K w = [H 3 O + ] [OH ‒ ] 14 Autoionization of water

15. Equilibrium Constant in Aqueous Solutions For any aqueous solution: K w = [H 3 O + ] [OH ‒ ] = 1.0 × 10 -14 at 25  C Remember that K w is always constant at a given temperature. However, there can be infinite possibilities of the equilibrium positions for aqueous solution. – When [H 3 O + ] = [OH ‒ ] ; the solution is neutral. – When [H 3 O + ] > [OH ‒ ] ; the solution is acidic. – When [H 3 O + ] < [OH ‒ ] ; the solution is basic. 15

16. Equilibrium Constant in Aqueous Solutions 16

17. Equilibrium Constant in Aqueous Solutions Exercise: Calculate [OH ‒ ] in a solution in which the concentration of protons is 0.0012 M at 25  C. Is the solution acidic, basic or neutral? At 25  C, K w is always equal to 1 × 10 -14. K w = [H 3 O + ] [OH ‒ ] 1 × 10 -14 = (0.0012) [OH ‒ ] [OH ‒ ] = 1 × 10 -14 / 0.0012 = 8.3 × 10 -12 M The solution is acidic. 17

18. The pH Scale The pH scale is a convenient way to express the acidity and basicity of a solution (the concentration of H 3 O + ions). The pH of a solution is defined as: pH = -log [H 3 O + ] or pH = -log [H + ] Example: Two solutions with hydronium ion concentrations of 1.0 × 10 -4 M and 5.0 × 10 -9 M. Find their pH values. The first solution; pH = - log (1.0 × 10 -4 ) = 4.00 The second solution; pH = - log (5.0 × 10 -9 ) = 8.30 18 2 sig. fig. 2 decimal places The number of sig. fig in the log is equal to the number of decimal places in the answer

19. pH scale Example 1 Calculate the pH of a 2.0 x 10 -3 M solution of HCl.

20. pH scale Solution [H + ] = 2.0 x 10 -3 M pH = − log 10 [H + ] =− log 10 [2.0 x 10 -3 M ] = 3 - log 10 2.0 = 3 – 0.30 = 2.7

21. The pH Scale Exercise: Find the pH of a neutral solution. A neutral solution has [H 3 O + ] = [OH ‒ ]. Since K w = [H 3 O + ] [OH ‒ ] = 1.0 × 10 -14 then [H 3 O + ] = 1.0 × 10 -7 pH = - log (1.0 × 10 -7 ) = 7.00 21 NeutralHigher pH values ( > 7.00 ) Lower [H 3 O + ] Lower pH values ( < 7.00 ) Higher [H 3 O + ] Basic solutionAcidic solution

22. The pOH Scale 22 A pOH scale is analogous to the pH scale. It is defined as: pOH = - log [OH ‒ ] Also, [OH ‒ ] = 10 -pOH

23. Kw = [H + ] [OH - ] -log K w = - log [H + ][OH - ] = -log [H + ] – log [OH - ] pK w = pH + pOH

24. pOH scale PLEASE NOTE THAT AT 25˚C [H + ] x [OH - ] = 10 -14 pH + pOH = 14

25. The pOH Scale Recall that at 25  C: K w = 1.0 × 10 -14 = [H 3 O + ] [OH ‒ ] ‒ log ([H 3 O + ] [OH ‒ ]) = ‒ log (1.0 × 10 -14 ) ‒ log [H 3 O + ] ‒ log [OH ‒ ] = 14.00 pH + pOH = 14.00 25 pH 13.00 11.00 9.00 7.00 5.00 4.00 1.00

26. Example Calculate the pOH and the pH of a 5.0 x 10 -2 M solution of NaOH

27. Solution Method 1 [OH - ] = 5.0 x10 -2 M pOH = - log(5.0 x10 -2 ) = 2 – log 5.0 = 1.3

28. pOH scale Solution continued Since pH + 1.3 = 14.00 pH = 12.7 pH + pOH = 14

29. pOH scale Solution Method 2 Since [H+] = 1.0 x 10 -14 = 2.0 x 10 -13 5.0 x 10 -12 pH = - log (2.0 x 10 -13 ) = 13 – log 2.0 = 13 – 0.30 = 12.7 [H + ] x [OH - ] = 10 -14

30. The pOH Scale Exercise: What is the pOH of a solution that has a hydroxide ion concentration of 4.3 x 10 -2 M? Exercise: What is the hydroxide ion concentration of a solution with pOH 8.35? 30